Question

how many terms of the taylor series for \\( \tan ^ { - 1 } x \\) would you have to use to evaluate each term on the right side of the equation \\( \pi = 48 \tan ^ { - 1 } \frac { 1 } { 18 } + 32 \tan ^ { - 1 } \frac { 1 } { 57 } - 20 \tan ^ { - 1 } \frac { 1 } { 239 } \\) with an error of magnitude less than \\( 10 ^ { - 6 } \\)?
you would have to use \\( \square \\) terms.

Explanation:

Step1: Recall the Taylor series for \(\tan^{-1}x\)

The Taylor series for \(\tan^{-1}x=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{2n + 1}x^{2n+1}\), and the error bound for the alternating series \(\sum_{n = 0}^{\infty}(-1)^{n}a_{n}\) (where \(a_{n+1}\leq a_{n}\) and \(\lim_{n
ightarrow\infty}a_{n}=0\)) is \(|R_{N}|\leq a_{N + 1}\). For \(\tan^{-1}x=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{2n+1}\), the error \(|R_{N}|\leq\frac{|x|^{2N + 3}}{2N+3}\)

Step2: Analyze the three terms \(48\tan^{-1}\frac{1}{18}\), \(32\tan^{-1}\frac{1}{57}\) and \(-20\tan^{-1}\frac{1}{239}\) separately

• For \(y_1 = 48\tan^{-1}\frac{1}{18}\): Let \(x=\frac{1}{18}\), we want \(48\times\frac{(\frac{1}{18})^{2N+3}}{2N + 3}<\frac{10^{-6}}{3}\) (since we divide the total error bound \(10^{-6}\) equally among the three terms). So \(\frac{48}{(2N + 3)}\times\frac{1}{18^{2N+3}}<\frac{10^{-6}}{3}\), or \(\frac{144}{(2N + 3)}\times\frac{1}{18^{2N+3}}<10^{-6}\)
• For \(y_2=32\tan^{-1}\frac{1}{57}\): Let \(x = \frac{1}{57}\), we want \(32\times\frac{(\frac{1}{57})^{2N+3}}{2N+3}<\frac{10^{-6}}{3}\). So \(\frac{96}{(2N + 3)}\times\frac{1}{57^{2N+3}}<10^{-6}\)
• For \(y_3=- 20\tan^{-1}\frac{1}{239}\): Let \(x=\frac{1}{239}\), we want \(20\times\frac{(\frac{1}{239})^{2N+3}}{2N+3}<\frac{10^{-6}}{3}\). So \(\frac{60}{(2N + 3)}\times\frac{1}{239^{2N+3}}<10^{-6}\)

We use the most restrictive condition. Consider the term with the largest \(x\) value (since the error formula \(\frac{|x|^{2N+3}}{2N + 3}\) is an increasing function of \(|x|\) for \(|x|\in(0,1)\)). The largest \(x\) among \(\frac{1}{18},\frac{1}{57},\frac{1}{239}\) is \(x=\frac{1}{18}\)

We solve \(\frac{48}{2N+3}\times\frac{1}{18^{2N + 3}}<\frac{10^{-6}}{3}\)

Let's try \(N = 3\)

For \(N = 3\), \(\frac{48}{2\times3+3}\times\frac{1}{18^{2\times3+3}}=\frac{48}{9}\times\frac{1}{18^{9}}=\frac{16}{3}\times\frac{1}{18^{9}}\approx\frac{16}{3}\times1.6\times10^{-11}<10^{-6}\)

Answer:

\(3\)